Hi to all.
Imagine that we have a supply cable that feeds the main busbars of a switchboard.
Lets also say that from these busbars depart 3 secondary supply cables that feed three motors, say A,B and C.
Lets assume that Ia=10 A, Ib=20 A and Ic=30A are the nominal currents of the motors.
Could we say that the supply cable is charged with 10+20+30=60A and start the calculations about the CSA with this number???
These are the rms values of the motor currents that are vectors and so the correct sum should be I = Ia+Ib+Ic but with the currents as vectors and not as rms values.
Is there any possible explanation why I should just add rms values in order to find the current of the supply cable?
I have heard a fellow engineer saying this thing but I do not agree and I would like to discuss it in here in case there is some simple explanation that I ignore.
Thanks!

The current loading will be cummulative as more and more load is designed/added for/to a DB. You have diversity in your favour for sub-main cables but VD, starting currents, EFLI and other factors can work against you.

Vectors are for mechanics - phasors have been the preferred term in most text books. Phasors rotate whilst vectors are generally static.

Anyway to address your question - as you are probably aware the motor loads are inductive and the current will be displaced from the voltage (assuming this is our reference) by an angle - the 'phase angle'. The cosine of this angle is equal to the power factor of the motor if we can assume sinusoidal conditions.

Each motor will have individual characteristics and so the power factors may be different for each motor and this will also vary with loading.

If the motors are all around the same size and have similar characteristics then summing the currents arithmetically will not introduce much error. However, strictly speaking the currents should be summed as phasors (vectors) so as to take account of any difference in angle from the reference.

Originally posted by: Apostolos1983Could we say that the supply cable is charged with 10+20+30=60A and start the calculations about the CSA with this number?

These currents are not additive for A.C. currents, they are the resultant of the true currents and the imaginary currents, the magnitude of these depends upon the power factor - Cos phi. As a starting point, see the "j notation" for single phase application. Z = R + jX etc., (Impedance, Resistance and Reactance).

Is there any possible explanation why I should just add rms values in order to find the current of the supply cable?

The currents can't be added as explained above. Apart from the thermal running currents for cable sizing, the voltage drop on starting for the motors, is very important and requires to be considered; and whether staggered starting is possible, to alleviate summation of these starting currents etc. etc. The power factor and torque output on D.O.L. S.C. motor starting is very low; if the voltage drop is excessive i.e. >30%, the motor/s may not start at all.

BTW "Fellow engineer" implies being non-electrical due to the lack of understanding shown.

No can add just values. These are in deed imaginary numbers, or phasors as GeoffBlackwell said.

I just wanted to know if there was any simple explanation that could lead in such a "simplification".

and in your first post

I have heard a fellow engineer saying this thing but I do not agree and I would like to discuss it in here in case there is some simple explanation that I ignore.

A word of caution before you tell all of your fellow engineers that they are wrong .

When you are designing an installation many assumptions may have to be made. The design can involve a great deal of time and someone has to pay for this. So if the process can be speeded up by making 'engineering judgements' this will be done provided the overall design is not compromised.

Take the example of selecting cable sizes - the solution i.e. the optimium cable size - will evolve after considering various factors. These factors will doubtless contain simplyfing assumptions (the adiabatic equation being a good example ).

The end result of this process is a cable size, but you must bear in mind that cables are not made in an infinite range of sizes, they are made to preffered sizes (normally based on some mathematics) such as: 10.0; 16.0; 25.0; 35.0 and so on usually up to 400.0 (all sizes in square millimetres).

So there is little point in directing too much effort to a cable size if the solution is going to be finally determined by a particular size availability. So if initial calculations suggest a 90.0mm2 cable there is little point in trying to reduce it as the size below is 70.0mm2 - a 95.0mm2 is going to be required. If the initial calculations suggest a 98mm2 cable there may be some point in trying to improve the accuracy of the calculations by using a more complex approach as this may allow the use of a 95mm2 cable.

If your fellow engineers are referring to a group of motors with similar sizes and characteristics their simplification may be acceptable. If not a more accurate approach is needed.

Sums for zeeper
As Jaymack indicated above - the design of motor circuits can be complex - so simple answers will not suffice for a 'real world' design.

However, the principles can often be extracted and simplified to allow learning to progress. So with that in mind a simple example may help.

Consider two loads - both have inductive characteristics (they could be motors, but lets not go there ).

Load A has a full load current of 10A and a full load power factor of 0.8 lagging.

Load B has a full load current of 10A and a full load power factor of 0.6 lagging.

These loads are connected to a remote distribution board and your task is to determine the magnitude of the current that will flow in the distribution circuit (sub-main) supplying the remote DB - when both loads are demanding their full load.

If you think the answer is 20A you have some way to go before you achieve enlightenment .

Both loads have some inductance and this causes the power factor to decrease from the value of 1.0 associated with resistive loads. However they have different power factors which implies that the inductive elements of each load are not the same. This means that the angle between voltage and current for each load will differ.

Now when considering this problem we need to establish a reference so that we can determine the phase difference or phase angle differences for each current.

The supply voltage is common to both loads at the origin of the system so the voltage phasor is our common reference. The angles between voltage and currents are referenced to this voltage.

For sinusoidal waves the power factor is numerically equal to the cosine of phase angle between voltage and current.

So for
Load A: Phase Angle = arccos(0.8) = 37 degrees

Load B: Phase Angle = arccos(0.6) = 53 degrees

So if we were to sketch the phasor diagram using the voltage as a horizontal reference - current A would have magnitude 10 but lagging phase angle 37 degrees and current B would have magnitude 10 and lagging phase angle 53 degrees.

The current in the distribution circuit will be the phasor sum of these two current and it will be less than 20A .

Correct example, taking into account an even distribution of the loads between the three phases. But let us assume (because we engineers never rest ) that the one of the two loads is single phase and not three phase. What happens in this case??? One of the three phases shall be charged a little bit more than the others isn't it?

The whole previous analysis of your example was based on the single phase equivalent circuit where we assume that the current in all phases has the same magnitute and from phase to phase the currents differ only by a 120 degree angle.

What happens if one or more of the loads are single phase? Obviously you cannot use the single phase equivalent circuit since at least one phase is different now.

(Is there any chance I do not remember the electrical circuit analysis well? In this case ignore my small analysis...)

Of course.
Each and every time that you add loads with different power factors, the final power factor is different from the previous ones. Only when you add loads with the same PF you get a final load with the same PF.

You lost me at the converting polar. They not teach that on my course lol.

Think I need to go back to school, I managed to get to 45 degrees using my fingers, but when I draw the phasor the result keeps coming back as 14A. I think I may have over simplified or missed a step. But I was surprised that a lag of 44.6 was only worth 0.2A

I think I'll stick with the current rating on the motor ID plates and worry about the start up voltdrop instead.

That is correct. In fact this is part of the procedure of converting an imaginary number into polar mode.
The angle is also given by -arctan (Im / Re) but I do not remember it by heart....(too much time since I last used it)